
Rotational Dynamics is the study of how forces cause rotational motion.
Just as Newton’s Second Law describes translational motion:
∑F=ma\sum F = ma∑F=ma
Rotational Dynamics uses the rotational equivalent:
∑τ=Iα\sum \tau = I\alpha∑τ=Iα
where:
This equation is one of the most important relationships in AP Physics C Mechanics.
A force can produce:
Whether a force causes rotation depends on where the force is applied relative to the axis of rotation.
For example:
The quantity that measures rotational effectiveness is torque.
Torque is the rotational equivalent of force.
It measures the tendency of a force to cause rotation.
Vector form:
τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}τ=r×F
where:
τ=rFsinθ\tau = rF\sin\thetaτ=rFsinθ
where:
Occurs when:
θ=90∘\theta=90^\circθ=90∘
Then:
τ=rF\tau=rFτ=rF
The force is perpendicular to the radius.
Occurs when:
θ=0∘\theta=0^\circθ=0∘
or
θ=180∘\theta=180^\circθ=180∘
The force acts through the axis.
Then:
τ=0\tau=0τ=0
Torque is a vector.
Its direction is determined by the right-hand rule.
Common convention:
The Moment of Inertia measures an object’s resistance to angular acceleration.
It is the rotational equivalent of mass.
Symbol:
III
Units:
kg⋅m2\text{kg}\cdot\text{m}^2kg⋅m2
For a point mass:
I=mr2I=mr^2I=mr2
where:
Mass farther from the axis contributes more strongly because of the squared distance.
For extended bodies:
I=∫r2 dmI=\int r^2\,dmI=∫r2dm
This calculus expression is used to derive moments of inertia for rods, disks, rings, and spheres.
I=MR2I=MR^2I=MR2
I=12MR2I=\frac12 MR^2I=21MR2
I=12MR2I=\frac12 MR^2I=21MR2
I=25MR2I=\frac25 MR^2I=52MR2
I=112ML2I=\frac1{12}ML^2I=121ML2
I=13ML2I=\frac13 ML^2I=31ML2
The rotational form of Newton’s Second Law is:
∑τ=Iα\sum\tau = I\alpha∑τ=Iα
This equation plays the same role in rotational motion that:
∑F=ma\sum F=ma∑F=ma
plays in translational motion.
| Linear Motion | Rotational Motion |
|---|---|
| Force FFF | Torque τ\tauτ |
| Mass mmm | Moment of Inertia III |
| Acceleration aaa | Angular Acceleration α\alphaα |
| Momentum ppp | Angular Momentum LLL |
Understanding this analogy simplifies rotational mechanics.
A disk has:
I=2 kg⋅m2I=2\,\text{kg}\cdot\text{m}^2I=2kg⋅m2
A torque of:
8 N⋅m8\,\text{N}\cdot\text{m}8N⋅m
is applied.
Find the angular acceleration.
Using:
∑τ=Iα\sum\tau=I\alpha∑τ=Iα 8=(2)α8=(2)\alpha8=(2)α α=4 rad/s2\alpha=4\,\text{rad/s}^2α=4rad/s2
An object is in rotational equilibrium when:
∑τ=0\sum\tau=0∑τ=0
In this situation:
α=0\alpha=0α=0
The object:
Many real objects both translate and rotate.
Examples:
In such problems:
∑F=ma\sum F=ma∑F=ma
and
∑τ=Iα\sum\tau=I\alpha∑τ=Iα
must often be applied simultaneously.
Using force instead of perpendicular force.
Only the perpendicular component contributes to torque.
Ignoring the distance from the axis.
Torque depends on the lever arm:
τ=rFsinθ\tau=rF\sin\thetaτ=rFsinθ
Confusing mass and moment of inertia.
Objects with identical masses can have different moments of inertia depending on how the mass is distributed.
Torque:
τ=rFsinθ\tau=rF\sin\thetaτ=rFsinθ
Moment of inertia:
I=mr2I=mr^2I=mr2
for a point mass.
Rotational Newton’s Second Law:
∑τ=Iα\sum\tau=I\alpha∑τ=Iα
Key ideas:
Rotational Dynamics provides the foundation for understanding rotating systems and leads directly to the study of rotational energy, rolling motion, and angular momentum in AP Physics C Mechanics.
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